Minimun depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root){
            return 0;
        }
        if (!root->left && !root->right){
            return 1;
        }
        if (root->left && !root->right){
            return minDepth(root->left) + 1;
        }
        if (root->right && !root->left){
            return minDepth(root->right) + 1;
        }
        return min(minDepth(root->left) + 1, minDepth(root->right) + 1);
    }
};

But....

You know the above algorithm is not the optimical. Pre-order complexity is O(n).

we can find more effective algorithm using BFS on complexity O(k) k is the minimun depth of tree

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    struct Node{
        TreeNode * p;
        int layer;

        Node(TreeNode * p, int layer){
            this->p = p;
            this->layer = layer;
        }
    };
    int minDepth(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root){
            return 0;
        }
        queue<Node > q;
        q.push(Node(root,1));

        while(!q.empty()){
            Node node = q.front();
            q.pop();

            if (!node.p->left && !node.p->right){
                return node.layer;
            }

            if (node.p->left){
                q.push(Node(node.p->left,node.layer+1));
            }
            if (node.p->right){
                q.push(Node(node.p->right,node.layer+1));
            }
        }

    }
};

原文链接: https://www.cnblogs.com/kwill/archive/2013/02/14/2911077.html

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